Question

If x= a(sin theta+ cos theta)
Y= b(sin theta - cos theta). ,then x^2/a^2+y^2/b2=???

Answer 1

x/a=sintheta + costheta
y/b=sin(theta)-cos(theta)
(x/a)²+(y/b)²=(sintheta +costheta)² + ( sintheta- costheta)²
sin²theta + cos²theta +2sintheta*costheta +sin²theta + cos²theta -2sintheta*costheta
2(sin²theta + cos²theta)
2*1=2

Answer 2

x = a( sinA + cosA )


$\underline{Square \: \: o n \: \: both \: \: sides, }$


➡ x² = a²( sinA + cosA )²

$= > x^{2} = a^{2} ( \bold{ sin^{2} A + cos ^{2} A} + 2sinAcosA ) \: \: \: \\ \\ = > \frac{ {x}^{2} }{ {a}^{2} } = 1 + 2sinAcosA \: \: \: \: \: \: -----: ( 1 )$





y = b( sinA - cosA )



$\underline{Square \: \: on \: \: both \: \: sides, }$


➡ y² = b²( sinA - cosA )²

$= > y^{2} = b^{2} ( \bold{sin^{2} A + cos^{2} A } - 2sinAcosA ) \\ \\ = > \frac{ {y}^{2} }{ {b}^{2} } = 1 - 2sinAcosA \: \: \: - - - - - - - (2)$








Hence,


$\frac{{x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }$



Putting values from ( i ) & ( ii ) ,


1 + 2sinAcosA + 1 - 2sinAcosA

1 + 1 + 2sinAcosA - 2sinAcosA

2








Answer : 2